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A plot of ln[ A ] versus t for a first-order reaction is a straight line with a slope of − k and an intercept of ln[ A ] 0 . If a set of rate data are plotted in this fashion but do not result in a straight line, the reaction is not first order in A .
![A graph is shown with the label “Time ( h )” on the x-axis and “l n [ H subscript 2 O subscript 2 ]” on the y-axis. The x-axis shows markings at 6, 12, 18, and 24 hours. The vertical axis shows markings at negative 3, negative 2, negative 1, and 0. A decreasing linear trend line is drawn through five points represented at the coordinates (0, 0), (6, negative 0.693), (12, negative 1.386), (18, negative 2.079), and (24, negative 2.772).](/ocw/mirror/col11760/m51101/CNX_Chem_12_04_FrstOKin.jpg)
| Trial | Time (h) | [H 2 O 2 ] ( M ) | ln[H 2 O 2 ] |
|---|---|---|---|
| 1 | 0 | 1.000 | 0.0 |
| 2 | 6.00 | 0.500 | −0.693 |
| 3 | 12.00 | 0.250 | −1.386 |
| 4 | 18.00 | 0.125 | −2.079 |
| 5 | 24.00 | 0.0625 | −2.772 |
The plot of ln[H 2 O 2 ] versus time is linear, thus we have verified that the reaction may be described by a first-order rate law.
The rate constant for a first-order reaction is equal to the negative of the slope of the plot of ln[H 2 O 2 ] versus time where:
In order to determine the slope of the line, we need two values of ln[H 2 O 2 ] at different values of t (one near each end of the line is preferable). For example, the value of ln[H 2 O 2 ] when t is 6.00 h is −0.693; the value when t = 12.00 h is −1.386:
| Trial | Time (s) | [ A ] |
|---|---|---|
| 1 | 4.0 | 0.220 |
| 2 | 8.0 | 0.144 |
| 3 | 12.0 | 0.110 |
| 4 | 16.0 | 0.088 |
| 5 | 20.0 | 0.074 |
The plot of ln[
A ] vs.
t is not a straight line. The equation is not first order:
![A graph, labeled above as “l n [ A ] vs. Time” is shown. The x-axis is labeled, “Time ( s )” and the y-axis is labeled, “l n [ A ].” The x-axis shows markings at 5, 10, 15, 20, and 25 hours. The y-axis shows markings at negative 3, negative 2, negative 1, and 0. A slight curve is drawn connecting five points at coordinates of approximately (4, negative 1.5), (8, negative 2), (12, negative 2.2), (16, negative 2.4), and (20, negative 2.6).](/ocw/mirror/col11760/m51101/CNX_Chem_12_04_CYL1_img.jpg)
The equations that relate the concentrations of reactants and the rate constant of second-order reactions are fairly complicated. We will limit ourselves to the simplest second-order reactions, namely, those with rates that are dependent upon just one reactant’s concentration and described by the differential rate law:
For these second-order reactions, the integrated rate law is:
where the terms in the equation have their usual meanings as defined earlier.
The reaction is second order with a rate constant equal to 5.76 10 −2 L/mol/min under certain conditions. If the initial concentration of butadiene is 0.200 M , what is the concentration remaining after 10.0 min?
We know three variables in this equation: [ A ] 0 = 0.200 mol/L, k = 5.76 10 −2 L/mol/min, and t = 10.0 min. Therefore, we can solve for [ A ], the fourth variable:
Therefore 0.179 mol/L of butadiene remain at the end of 10.0 min, compared to the 0.200 mol/L that was originally present.
0.0196 mol/L
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